A pair of straight lines is represented by 2y^2 +5xy -3x^2 = 02y2+5xy3x2=0. Another line having an equation x+y = kx+y=k. These lines form a triangle whose centroid is (1/18,11/18)(118,1118). Find the value of k?

1 Answer
Sep 20, 2017

k=1k=1

Explanation:

2y^2+5xy-3x^2=(2 y-x) (3 x + y) = 02y2+5xy3x2=(2yx)(3x+y)=0

so the three lines are

{(2 y-x=0),(3 x + y=0),(x+y=k):}

and given

{(2 y_1-x_1=0),(3 x_1 + y_1=0),(3x_2+y_2=0),(x_2+y_2=k),(x_3+y_3=k),(2y_3-x_3=0),((x_1+x_2+x_3)/3=1/18),((y_1+y_2+y_3)/3=11/18):}

where

p_1 = (x_1,y_1)
p_2=(x_2,y_2)
p_3=(x_3,y_3)

are the intersection points or the vertices.

Solving for x_1,x_2,x_3,y_1,y_2,y_3,k we obtain

x_1 = 0, x_2 = -1/2, x_3 = 2/3, y_1 = 0, y_2 = 3/2, y_3 = 1/3, k = 1

and then k = 1